This is part of a question(2.9.3) from Patrick Morandi's "Field and Galois Theory" I am just plain stuck on this fact:

$F(\sqrt{a})/F$ is a field extension where $a\in F-F^2$ and $F$ does not contain a primitive fourth root of unity. If $a$ is a sum of two squares in $F$, prove that $N_{F(\sqrt{a})/F}(\alpha)=-1$ for some $\alpha\in F(\sqrt{a})$ and that $N_{F(\sqrt{a})/F}(\alpha)=a$ for some $\alpha\in F(\sqrt{a})$. Hence, show that $N_{F(\sqrt{a})/F}(b)\equiv a \mod F^{*2}$ for some $b\in F(\sqrt{a})$.

I know that $\text{Gal}(F(\sqrt{a})/F=\{1,\sigma\}$ where $\sigma:\sqrt{a}\mapsto-\sqrt{a}$. Also, $\{1,\sqrt{a}\}$ is a basis for $F(\sqrt{a})$ over $F$, so the norm of any element $c_1+c_2\sqrt{a}\in F(\sqrt{a})$ is just $\left(c_1+c_2\sqrt{a}\right)\left(c_1-c_2\sqrt{a}\right)=c_1^2-c_2^2a$.

The fact that the norm map is surjective when $F$ is a finite field gives us the first two facts right away. But it has nothing to do with $a$ being a sum of squares, or that $F$ has no fourth root of unity, and of course it only works for finite fields.

Also, if you compute $\min(\sqrt{a-1},F)=x^2-a-1$, then $N_{F(\sqrt{a})/F}=(-1)^n(-1)^{\frac{n}{2}}=-1$ since $n=[F(\sqrt{a}):F]=2$. but this also seems wrong because it still has nothing to do with the premises.

How might I proceed here?

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    $\begingroup$ $\sqrt{a-1}$ is not in your field $\endgroup$ – reuns Aug 10 at 3:51

$$a = u^2+v^2, \qquad N_{F(\sqrt{a})/F}(\frac{u+\sqrt{a}}{v}) =\frac{u+\sqrt{a}}{v}\frac{u-\sqrt{a}}{v}= \frac{u^2-a}{v^2} =-1,\\ \qquad N_{F(\sqrt{a})/F}(\sqrt{a}\frac{u+\sqrt{a}}{v}) = - N_{F(\sqrt{a})/F}(\sqrt{a}) = a$$


Your notations are a bit shaky: $L$ is not defined, although it should naturally mean $F(\sqrt a)$ ; in the first question, the two $\alpha$ should not be the same. This being said, $a$ is a sum of squares in $F$ iff $a$ is a norm from $F(\sqrt {-1})$ (which is a quadratic extension of $F$ by hypothesis), and the latter condition is plainly equivalent to $-1$ being a norm from $F(\sqrt a)$. Since $-a$ is already a norm from $F(\sqrt a)$, so will be $a$.


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